Sum of floors inequality

Let $x=A+a,6x=B+b,10x=C+c,15x=D+d,30x=E+e$ where big letter is the integer part and small letter is the decimal part.

(1) If $a<{1\over30}$ we have $$b=6a\implies B=6A$$ $$c=10a\implies C=10A$$ and so on $D=15A,E=30A$. So $A+E=B+C+D$.

(2) If ${1\over30}\leq a<{1\over 15}$ then we have $B=6A,C=10A,D=15A$ $$e=30a-1\implies E=30A+1$$ so $A+E>B+C+D$

(3) If ${1\over 15}\leq a<{1\over 10}$ then we have $B=6A,C=10A$ $$d=15a-1\implies D=15A+1$$ $$e=30a-2\implies E=30A+2$$ so $A+E>B+C+D$

Can you finish off the remaining two cases?

I have a quite tedious argument that this inequality holds:

Firstly, by taking $x \pmod 1$ we may assume $x \in [0,1)$. Let $k$ be the greatest integer, such that $\frac{k}{30} \leqslant x$ (so $k = \lfloor 30x \rfloor$). Then we would want to prove that

$$\lfloor 15x \rfloor = \bigg \lfloor \frac{k}{2} \bigg \rfloor.$$

Firstly, $\lfloor 15x \rfloor \geqslant \big\lfloor \frac{k}{2} \big\rfloor$, as $15x \geqslant \frac{k}{2}$. Suppose $\lfloor 15x \rfloor > \big\lfloor \frac{k}{2} \big\rfloor$, then we can choose an integer $l$ such that $15x \geqslant l > \frac{k}{2}$. But that would mean $\frac{2l}{30} \leqslant x$ and $2l > k$, which is a contradiction with a definition of $k$.

Using the same methods, we can transform our inequality to

$$ k+ \bigg\lfloor \frac{k}{30} \bigg\rfloor \geqslant \bigg\lfloor \frac{k}{2} \bigg\rfloor + \bigg\lfloor \frac{k}{3} \bigg\rfloor + \bigg\lfloor \frac{k}{5} \bigg\rfloor $$

where $k$ is an integer between 0 and 29. That's $30$ cases to check, but they're all simple so it can be done quite quickly.

EDIT We can also transform RHS using $\lfloor a \rfloor + \lfloor b \rfloor \leqslant \lfloor a+b \rfloor$. Then we will obtain inequality $k + \big\lfloor \frac{k}{30} \big\rfloor \geqslant \big\lfloor \frac{31k}{30} \big\rfloor$, which is not true in general case, but it holds for $k=0,\ 1,\ldots,29$.